Introduction to

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Introduction to

Bioinformatics

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Introduction to Bioinformatics.

LECTURE 8: Whole genome comparisons

* Chapter 8: Welcome to the Hotel Chlamydia

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Introduction to BioinformaticsLECTURE 8: WHOLE GENOME COMPARISONS

8.1 Uninvited guests

* Symbionts: organisms that live together in a beneficial relation

* E. coli and Human: receives nutrients, gives vitamin K

* Numerous examples in Nature: flowers and bees, tick- bird and rhinoceros, pea aphid and Buchnera, mitochondria and Eukaryotes

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8.1 Uninvited guests

* Some symbionts have moved permanently into the cells of the host

* They have become entirely dependent on the host to provide them with nutrients, oxygen, specific proteins …

* In the process they have lost many genes necessary to produce such products themselves

* As a result, intracellular obligate symbionts have the smallest genomes – both in total size as in number of genes

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Introduction to Bioinformatics8.1 – UNINVITED GUESTS

Chlamydia trachomatis

* Chlamydia trachomatis is an intracellular symbiont that gives no benefit to the host : it is a parasite

* C. trachomatis is the most common sexually transmitted disease with +/- 3M new infections per annum in the USA

* It has lost the ability to produce many biochemical products and must live in specific cells in the human (hence the characterisation of: obligate endo-symbiont)

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Chlamydia trachomatis

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Chlamydia pneunomia

* Chlamydia pneumonia is a related bacterial parasite of the human respiratory tract : it causes pneumonia and bronchitis

* Like C. trachomatis it has a very small genome ~ 1 Mb

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Human respiratory tractChlamydia pneunomia

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Chlamydia pneunomia

Phylogenetic analysis of the parasitic lifestyle of Chlamydia shows that it dates back to 700 Myrs with the emergence of

Eukaryotes

* This is the same date as the pure symbiontic lifestyle of mitochondria

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Whole genome comparisons

* In this lecture we study the problems involved with the comparisons of entire genomes

* Because of its very small genome Chlamydia are a perfect case study

* Moreover, Chlamydia shows a high conservation of the order of the genes and virtually no horizontal gene transfer.

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Hotel Chlamydia

‘Hotel’ Chlamydia is not so much that we are a living hotel for the Chlamydia (which is true), but that genes are guest in hotel Chlamydia – guests that come in, reshuffle rooms, move out, pass along …

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Introduction to Bioinformatics LECTURE 8: WHOLE GENOME COMPARISONS

8.2 Patterns of genome evolution

* Genome comparison looks at the differences between the entire set of genes between two organisms

* This provides insight in evolution and function of genes

* Single nucleotide polymorphisms form the bulk of the genetic variability

* Also rearrangement and shuffling of genes: inversion, duplication, translocation

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8.2 Patterns of genome evolution

* Often translocation between two organisms: horizontal gene transfer

* Some 20% of E. coli‘s genes derive from horizontal transfer

* Chromosomes can break apart and or stick together

* Whole genomes can be duplicated → polyploid individuals

* This is the basis for new functions as these extra genes are free to evolve

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8.3 Beanbag genomics

* Genome = beanbag of genes + junk-DNA

* Comparison of whole genomes is more than comparing individual genes

* inversions, transpositions, duplications, deletions, chromosomal rearrangements

* Therefore an alignment of entire genomes will not work

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8.3 - BEANBAG GENOMICS : Comparison of two genomes

Basic mechanisms of gene evolution

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ACTTTTTGGGAC

inversion transposition

GGG

duplication

GGG

ACTTTTTGG TATATA CATGTAGTAC AAATAATCG AACCCCCGGAC GGG

TATATA CATGTAGTAC AAATAATCG AACCCCCG

deletion

TATATA

an alignment of entire genome will not work !

Therefore we have to break the problem in smaller pieces and build it back up …

… with multiple single-gene analysis

gene evolution

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8.3 - BEANBAG GENOMICS : Comparison of two chromosomes

AAATAATCG AACCCCCGACTTTTTGG TATATA CATGTAGTACGAC GGG

chromosome evolution

CATGTAGTACAAATAATAACCCCCG ACTTTTTGGTATATA CGGACGGG

Splitting into new chromosomesReshuffling of genes over the chromosomes

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* STEP 1: Find which genes are present in both

* STEP 2: use ORF-finder with threshold of 100 codons

* EXAMPLE: Chlamydia trachomatis and C. pneumonia :

--- Organism --------- size (nt) --- ORFs ---

C. trachomatis 1 042 519 916

C. pneumonia 1 229 853 1048E. coli - 5000


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* Intracellular symbionts like CT and CP have lost many genes: they parasite on their host and ‘steal’ the gene-products

* CT lives in urinary tracts and CP lives in respiratory tracts

* The differences in their genomes tells us something about the function of their retained genes

* What are suitable algorithmic methods for comparing lost and gained genes in Chlamydia?

--- Organism --------- size (nt) --- ORFs ---

C. trachomatis 1 042 519 916

C. pneumonia 1 229 853 1048

E. coli - 5000

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Similarity on a genomic scale

Similarity between pairs of genes informs about:

* Blocks of conserved gene order

* Changes in size of gene families

* Nucleotide substitutions between orthologous genes

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Central idea for genomic comparison

Define the similarity scores between genomes as:

* Nucleotide sequences of all genes found in both genomes

* Fill out a matrix with alignment scores between each possible pair of sequences

* For the Chlamydiae CT and CP this is a 1048x916 matrix

* Use Needleman-Wunsch or BLAST to compute similarity scores

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Identifying orthologous and paralogous genes

* Use genome similarity matrix to distinguish between paralogs and orthologs

* remember: homologs are genes that have a common ancestor, orthologs arise as homologs evolve in sister- species; paralogs arise from duplication and subsequent specialisation

* Result of evolution of homologs and paralogs: no one-to- one relationship, but (many/one)-to-(many/one)

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Reciprocal similarity

* Recognition of orthologs: Best Reciprocal similarity Hits (BRHs)

* A pair of ORFs is a BRH if it is the best match between the two genomes (using alignment scores)

* possible: ORFs without BRH

* possible: ORF with ortholog in other species and a paralog

in the same species.

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EXAMPLE 8.1: Homology in Chlamydia

* Similarity matrix → BRH → orthologs

* With threshold = 100 codons we find 1964 ORFs (CT: 916 and CP: 1048)

* Among these 1964 ORFs are 728 ortholog pairs

* Also 126 pairs of paralogs (CT: 56, CP: 70)

* These paralogs are more similar to each other than to orthologs → result of duplication after the species split

* The remaining 13% (=253 ORFs) perhaps older paralogs that have been lost in the other species due to specialisation

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Identifying gene families

* Defining a gene family is a tricky thing; at some high level all genes are ‘family’

* The very first DNA-based organism, ancestral to all present living beings, had a set of genes.

* All (?) genes have derived from these ancestral genes through duplication and subsequent specialization.

* Genes that cooperate tend to move close together (Dawkins: like rowers in a rowing boat)

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Identifying gene families

* Practical solution: only consider genes that are > 50% similar; they are ‘closely’ related and probably have a similar ‘function’

* Method for finding ‘similar’ genes: clustering

* Draw-back: all clustering methods have some degree of arbitrariness

* Cluster both genomes simultaneously, then count #genes in each cluster (=gene family).

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Identify gene families with Hierarchical Clustering

* input: genome similarity matrix d

* method: cluster d with NJ- or UPGMA-algorithm to group the genes in families

* This is called Hierarchical Clustering (HC)

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Hierarchical Clustering on both Chlamidiae

Application of HC on Chlamydia reveals a large number of small gene families and a small number of large families

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Similar function of gene families

Largest gene families in C. trachomatis and C. pneumonia:

-------- CT ---- CP ------ Function ----------------------------12 12 ABC transporters6 15 G family outer membrane protein9 10 Function not known9 10 Function not known

ABC transporters are transmembrane proteins with binding sites on both sites : major role in transport in/out the cell

They are very old: they are (near) identical in all organisms

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ABC transporter · Hydrophobicity

ABC transport ATP-binding cassette

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Schematic of the E. coli vitamin B12 importer system.

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Snapshots of pore formation in the bilayer, with an applied field of 0.5 V/nm in the presence of 1 M NaCl

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Alternative approaches to finding orthologs

* Clustering of genes in families has some arbitrariness

* Ortholog genes are separated by a speciation event

* Thus, a phylogenetic tree is also a useful metaphor

* A phylogenetic tree is a better representation, but it is less amenable to an automated analysis

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8.4 Synteny

* In Section 8.3 the emphasis was on analysis of genes, here the emphasis is on chromosomes

* syn- = together, tenia = ribbon, band,

* synteny : the relative ordering of genes on the same chromosomes

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Introduction to Bioinformatics 8.4 – SYNTENY

Major mechanisms of reshuffling of synteny are: inversions and transpositions

Noise on synteny is caused by:insertions, duplications, and deletions

Blocks of synteny: long stretches of DNA where the relative ordering of orthologous genes is conserved

Synteny allows for annotation of non-coding sequences and identification of homologous intergenetic regions

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Cat on Human

Conserved synteny map

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Visualising Synteny

Dot-plot:

* x-axis = position on genome_1,

* y-axis = position on genome_2,

* For a homologous gene with genome_1-position x, and genome_2-position y: put a dot ‘*’ on (x,y)

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Chlamidia Synteny

The high level of conservation in Chlamidia is remarkable but typical for all intracellular symbionts

For instance Buchnera aphidicola, a intracellular symbiont in pea aphids, has retained synteny for 50 million years.

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Buchneria, a endosymbiont in pea aphids, has retained synteny for 50 million years.

pea aphid Buchneria aphidicola

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SYMBIOSIS BETWEEN PEA APHIDS AND BUCHNERA

Plant sap contains little protein and aphids cannot produce ten essential amino acids

The required amino acids come from their symbiotic friends, the bacterium Buchnera aphidicola.

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SYMBIOSIS BETWEEN PEA APHIDS AND BUCHNERA

The symbionts genome reflects this biosynthetic activity.

Buchnera aphidicola carries the two genes trpEG for tryptophan synthesis. Each bacterium contains three or four plasmids that contain four tandem repeats of these genes, resulting in 12 to 16 copies of trpEG.

Thus, the symbionts supply the host with the essential amino acids and receives free nutrients and shelter.

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The relation between pea aphids and Buchnera aphidicola is very old ….

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Endosymbionts and Synteny

Buchnera aphidicola has retained synteny for 50 million years.

IN GENERAL: The cloistered lifestyle of endo-symbiotic organisms shields them from viruses and other bacteria that may induce gene rearrangement

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Homologous intergenic regions and ‘phylogenetic footprinting’

Intergenic regions are not selected for → fast evolution

Non-protein coding regions of the genome that are conserved are suspicious: they may be RNA-coding or regulatory sequences

Using syntenic coding regions as anchors we can find intergenic regions that are highly conserved.

This is called: Genetic Footprinting

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ORF CT672

CP1950ORF CT671-CP1949 ORF CT673-CP1951

intergenic regions

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A metric for the syntenic distance

Two genomes can be formed by many smaller syntenic blocks rearranged by inversions or transpositions

Can we define a metric for the syntenic distance of these two genomes?

We are not interested in nucleotide differences but in the number of genomic rearrangements that separate the two species.

METRIC = smallest number of operations (=inversion or transposition) that transform one genome into the other

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EXAMPLE: Sorting by reversals

As an example let us consider only the case of inversions

“sorting by reversals” = minimum number of inversions to transform one genome into the other

Algorithm: given a permutation of N numbers find the shortest series of reversals that can sort the back into their original order

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EXAMPLE: Sorting by reversals

3 2 1 4 8 7 6 5 9 1 2 3 4 8 7 6 5 9

1 2 3 4 5 6 7 8 9

2 reversals → syntenic distance = 2

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Sorting by reversals

NOTE:

In practice we do not know the original genome, so we select either one of the two as ‘the’ standard

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SIMPLE REVERSAL ALGORITHM

STEP 1: designate one sequence as the standard s and the other as t

STEP 2: i=1, increase(i) until s(i) ≠ t(i) or i=length(t)

STEP 3: j=i; increase(j) until t(j) = s(i), reverse(t(i:j)

STEP 4: i=j+1; if i=length(t), stop, else goto STEP-2

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SIMPLE REVERSAL ALGORITHM

QUESTION:

Can this algorithm solve overlapping reversals?

REMARK :

Involving transpositions is even more complex …

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END of LECTURE 8

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